Practice: Vectors and Linear Combinations

(a) Total profit. $c^{\mathsf{T}}x=(4)(8)+(9)(3)+(2)(5)=32+27+10=69$.

(b) Effect of the change. $c^{\mathsf{T}}d=(4)(2)+(9)(-1)+(2)(0)=8-9=-1$. Moving along $d$ loses one unit of profit per step, so it makes the plan worse. The direction trades a unit of the £9 product for two of the £4 product, which is an £8-for-£9 exchange.

(c) A flat direction. Any nonzero $d^{\prime }$ with $c^{\mathsf{T}}{d}^{\prime }=0$ will do, for instance $d^{\prime }=(9,-4,0)$, since $(4)(9)+(9)(-4)+(2)(0)=36-36=0$. Such a direction is orthogonal to the cost vector: the plan changes but the profit does not.

What that means. A flat direction is how a linear program comes to have several optimal plans with the same value. If you can move along it and stay feasible, every point you pass through is worth exactly as much as the one you started from.

(a) Lengths. $\Vert x\Vert =\sqrt{6^2+8^2}=\sqrt{100}=10$ km. $\Vert y\Vert =\sqrt{3^2+4^2}=\sqrt{25}=5$ km.

(b) Unit vector. $x/\Vert x\Vert =(6/10,8/10)=(0.6,0.8)$. Check: ${0.6}^2+{0.8}^2=0.36+0.64=1$.

(c) Direction. Yes. $x=2y$, and multiplying by a positive number scales a vector without turning it, so both point the same way. Equivalently, both have the unit vector $(0.6,0.8)$.

Why the questions are separate. $x$ is twice as long as $y$ and points identically. Length answers how far; direction answers which way. Neither determines the other: two runs of equal length can point in opposite directions, and two runs along one bearing can differ in distance.

The error. The student has computed the entrywise product $(x_1{y}_1,x_2{y}_2)=(3,8)$, which is a different operation. The dot product sums those products into a single scalar:

Why it matters beyond arithmetic. Every later use of this notation expects one number. A cost applied to a plan is money; a constraint row applied to a point is a quantity to compare against a bound. A two-entry answer cannot be compared with a bound at all, so the error does not stay local. It makes the next step meaningless.

What the correct value reports. $11$ is positive, which does say the two vectors broadly agree in direction. So the student's conclusion happened to be right, reached by a route that does not support it: the sign of the scalar carries that information, not the signs of two entries that were never supposed to remain separate.

On 'amplifies'. The comparison of lengths is not meaningful here. The dot product is not a vector and has no length; if the student wants a length they want the norm, $\Vert x\Vert =\sqrt{17}$, which is a different question again.

What is being computed. The left-hand side is a dot product of the coefficient vector $a=(3,1,4)$ with the schedule $t$. It reports how much of the limited resource this schedule consumes. One number, in whatever units the $40$ is measured in.

Testing the schedule. $a^{\mathsf{T}}t=(3)(6)+(1)(5)+(4)(4)=18+5+16=39$. Since $39\le 40$, the schedule satisfies the restriction, with one unit of the resource to spare.

A change orthogonal to $a$. Suppose a proposed adjustment $d$ satisfies $a^{\mathsf{T}}d=0$. Then moving along $d$ changes the schedule without changing the resource consumed at all. The new schedule uses exactly the same $39$ units. Such a direction slides along the boundary rather than towards or away from it.

Why that is useful. If the resource is currently binding, an orthogonal direction is the only kind you can move along without violating it. That is exactly how an optimiser explores alternatives once a constraint is tight, and it is why the orthogonality test matters far beyond the geometry lesson that introduces it.

(a) What is right and wrong. Both norms are correct: $\Vert p\Vert =\sqrt{4+1}=\sqrt{5}$ and $\Vert q\Vert =\sqrt{64+16}=\sqrt{80}$, so $q$ really is the longer vector, four times as long, since $\sqrt{80}=4\sqrt{5}$. What does not follow is "a bigger vector points somewhere different". Length and direction are independent properties, and a positive multiple changes only the first.

(b) Unit vectors. $p/\Vert p\Vert =(2,1)/\sqrt{5}\approx (0.894,0.447)$ and $q/\Vert q\Vert =(8,4)/\sqrt{80}\approx (0.894,0.447)$. They are identical, so the two vectors point the same way. Dividing by the norm rescales without turning, which is why this comparison is the one that answers the direction question.

(c) The relationship. $q=4p$. As price structures they are proportional, the second good costs half the first in both, and differ only by a scale factor of four.